Engineering Mechanics: Statics

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Engineering Mechanics: Statics. Chapter 5: Equilibrium of a Rigid Body. Chapter Objectives. To develop the equations of equilibrium for a rigid body. To introduce the concept of the free-body diagram for a rigid body. To show how to solve rigid-body equilibrium problems using the

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Engineering Mechanics: Statics

Chapter 5: Equilibrium of a Rigid Body

Chapter Objectives
• To develop the equations of equilibrium for a rigid body.
• To introduce the concept of the free-body diagram for a rigid body.
• To show how to solve rigid-body equilibrium problems using the
• equations of equilibrium.
Chapter Outline
• Conditions for Rigid Equilibrium
• Free-Body Diagrams
• Equations of Equilibrium
• Two and Three-Force Members
• Equilibrium in Three Dimensions
• Equations of Equilibrium
• Constraints for a Rigid Body
5.1 Conditions for Rigid-Body Equilibrium
• Consider rigid body fixed in the x, y and z reference and is either at rest or moves with reference at constant velocity
• Two types of forces that act on it, the resultant internal force and the resultant external force
• Resultant internal force fi is caused by interactions with adjacent particles
5.1 Conditions for Rigid-Body Equilibrium
• Resultant external force Fi represents the effects of gravitational, electrical, magnetic, or contact forces between the ith particle and adjacent bodies or particles not included within the body
• Particle in equilibrium, apply Newton’s first law,

Fi + fi = 0

5.1 Conditions for Rigid-Body Equilibrium
• When equation of equilibrium is applied to each of the other particles of the body, similar equations will result
• Adding all these equations vectorially,

∑Fi + ∑fi = 0

• Summation of internal forces = 0 since internal forces between particles in the body occur in equal but opposite collinear pairs (Newton’s third law)
5.1 Conditions for Rigid-Body Equilibrium
• Only sum of external forces will remain
• Let ∑Fi = ∑F, ∑F = 0
• Consider moment of the forces acting on the ith particle about the arbitrary point O
• By the equilibrium equation and distributive law of vector cross product,

ri X (Fi + fi) = ri X Fi + ri X fi = 0

5.1 Conditions for Rigid-Body Equilibrium
• Similar equations can be written for other particles of the body
• Adding all these equations vectorially,

∑ri X Fi + ∑ri X fi = 0

• Second term = 0 since internal forces occur in equal but opposite collinear pairs
• Resultant moment of each pair of forces about point O is zero
• Using notation ∑MO = ∑ri X Fi,

∑MO = 0

5.1 Conditions for Rigid-Body Equilibrium
• Equations of Equilibrium for Rigid Body

∑F = 0

∑MO = 0

• A rigid body will remain in equilibrium provided the sum of all the external forces acting on the body = 0 and sum of moments of the external forces about a point = 0
• For proof of the equation of equilibrium,

- Assume body in equilibrium

5.1 Conditions for Rigid-Body Equilibrium

- Force system acting on the body satisfies the equations ∑F = 0 and ∑MO = 0

- Suppose additional force F’ is applied to the body

∑F + F’ = 0

∑MO + MO’= 0

where MO’is the moment of F’ about O

- Since ∑F = 0 and ∑MO = 0, we require F’ = 0 and MO’

- Additional force F’ is not required and equations ∑F = 0 and ∑MO = 0 are sufficient

5.2 Free-Body Diagrams
• FBD is the best method to represent all the known and unknown forces in a system
• FBD is a sketch of the outlined shape of the body, which represents it being isolated from its surroundings
• Necessary to show all the forces and couple moments that the surroundings exert on the body so that these effects can be accounted for when equations of equilibrium are applied
5.2 Free-Body Diagrams

Support Reactions

• If the support prevents the translation of a body in a given direction, then a force is developed on the body in that direction
• If rotation is prevented, a couple moment is exerted on the body
• Consider the three ways a horizontal member, beam is supported at the end

- roller, cylinder

- pin

- fixed support

5.2 Free-Body Diagrams

Support Reactions

Roller or cylinder

• Prevent the beam from translating in the vertical direction
• Roller can only exerts a force on the beam in the vertical direction
5.2 Free-Body Diagrams

Support Reactions

Pin

• The pin passes through a hold in the beam and two leaves that are fixed to the ground
• Prevents translation of the beam in any direction Φ
• The pin exerts a force F on the beam in this direction
5.2 Free-Body Diagrams

Support Reactions

Fixed Support

• This support prevents both translation and rotation of the beam
• A couple and moment must be developed on the beam at its point of connection
• Force is usually represented in x and y components
5.2 Free-Body Diagrams
• Cable exerts a force on the bracket
• Type 1 connections
• Rocker support for this bridge girder allows horizontal movements so that the bridge is free to expand and contract due to temperature
• Type 5 connections
5.2 Free-Body Diagrams
• Concrete Girder rest on the ledge that is assumed to act as a smooth contacting surface
• Type 6 connections
• Utility building is pin supported at the top of the column
• Type 8 connections
5.2 Free-Body Diagrams
• Floor beams of this building are welded together and thus form fixed connections
• Type 10 connections
5.2 Free-Body Diagrams

External and Internal Forces

• A rigid body is a composition of particles, both external and internal forces may act on it
• For FBD, internal forces act between particles which are contained within the boundary of the FBD, are not represented
• Particles outside this boundary exert external forces on the system and must be shown on FBD
• FBD for a system of connected bodies may be used for analysis
5.2 Free-Body Diagrams

Weight and Center of Gravity

• When a body is subjected to gravity, each particle has a specified weight
• For entire body, consider gravitational forces as a system of parallel forces acting on all particles within the boundary
• The system can be represented by a single resultant force, known as weight W of the body
• Location of the force application is known as the center of gravity
5.2 Free-Body Diagrams

Weight and Center of Gravity

• Center of gravity occurs at the geometric center or centroid for uniform body of homogenous material
• For non-homogenous bodies and usual shapes, the center of gravity will be given
5.2 Free-Body Diagrams

Idealized Models

• Needed to perform a correct force analysis of any object
• Careful selection of supports, material, behavior and dimensions for trusty results
• Complex cases may require developing several different models for analysis
5.2 Free-Body Diagrams

Idealized Models

• Consider a steel beam used to support the roof joists of a building
• For force analysis, reasonable to assume rigid body since small deflections occur when beam is loaded
• Bolted connection at A will allow for slight rotation when load is applied => use Pin
5.2 Free-Body Diagrams

Support at B offers no resistance to horizontal movement => use Roller

• Building code requirements used to specify the roof loading (calculations of the joist forces)
• Large roof loading forces account for extreme loading cases and for dynamic or vibration effects
• Weight is neglected when it is small compared to the load the beam supports
5.2 Free-Body Diagrams

Idealized Models

• Consider lift boom, supported by pin at A and hydraulic cylinder at BC (treat as weightless link)
• Assume rigid material with density known
• For design loading P, idealized model is used for force analysis
• Average dimensions used to specify the location of the loads and supports
5.2 Free-Body Diagrams

Procedure for Drawing a FBD

1. Draw Outlined Shape

• Imagine body to be isolated or cut free from its constraints
• Draw outline shape

2. Show All Forces and Couple Moments

• Identify all external forces and couple moments that act on the body
5.2 Free-Body Diagrams

Procedure for Drawing a FBD

• Usually due to

- reactions occurring at the supports or at points of contact with other body

- weight of the body

• To account for all the effects, trace over the boundary, noting each force and couple moment acting on it

3. Identify Each Loading and Give Dimensions

• Indicate dimensions for calculation of forces
5.2 Free-Body Diagrams

Procedure for Drawing a FBD

• Known forces and couple moments should be properly labeled with their magnitudes and directions
• Letters used to represent the magnitudes and direction angles of unknown forces and couple moments
• Establish x, y and coordinate system to identify unknowns
5.2 Free-Body Diagrams

Example 5.1

Draw the free-body diagram of the uniform

beam. The beam has a mass of 100kg.

5.2 Free-Body Diagrams

Solution

Free-Body Diagram

5.2 Free-Body Diagrams

Solution

• Support at A is a fixed wall
• Three forces acting on the beam at A denoted as Ax, Ay, Az, drawn in an arbitrary direction
• Unknown magnitudes of these vectors
• Assume sense of these vectors
• For uniform beam,

Weight, W = 100(9.81) = 981N

acting through beam’s center of gravity, 3m from A

5.2 Free-Body Diagrams

Example 5.2

Draw the free-body diagram of

the foot lever. The operator

applies a vertical force to the

pedal so that the spring is

stretched 40mm and the force

in the short link at B is 100N.

5.2 Free-Body Diagrams

Solution

• Lever loosely bolted to frame at A
• Rod at B pinned at its ends and acts as a short link
• For idealized model of the lever,
5.2 Free-Body Diagrams

Solution

• Free-Body Diagram
• Pin support at A exerts components Ax and Ay on the lever, each force with a known line of action but unknown magnitude
5.2 Free-Body Diagrams

Solution

• Link at B exerts a force 100N acting in the direction of the link
• Spring exerts a horizontal force on the lever

Fs = ks = 5N/mm(40mm) = 200N

• Operator’s shoe exert vertical force F on the pedal
• Compute the moments using the dimensions on the FBD
• Compute the sense by the equilibrium equations
5.2 Free-Body Diagrams

Example 5.3

Two smooth pipes, each

having a mass of 300kg, are

supported by the forks of the

tractor. Draw the free-body

diagrams for each pipe and

both pipes together.

5.2 Free-Body Diagrams

Solution

• For idealized models,
• Free-Body Diagram

of pipe A

5.2 Free-Body Diagrams

Solution

• For weight of pipe A, W = 300(9.81) = 2943N
• Assume all contacting surfaces are smooth, reactive forces T, F, R act in a direction normal to tangent at their surfaces of contact
• Free-Body Diagram at pipe B
5.2 Free-Body Diagrams

Solution

*Note: R represent the force of A on B, is equal

and opposite to R representing the force of B on A

• Contact force R is considered an internal force, not shown on FBD
• Free-Body Diagram of both pipes
5.2 Free-Body Diagrams

Example 5.4

Draw the free-body diagram

of the unloaded platform that

is suspended off the edge of

the oil rig. The platform has a

mass of 200kg.

5.2 Free-Body Diagrams

Solution

• Idealized model considered in 2D because by observation, loading and the dimensions are all symmetrical about a vertical plane passing through the center
• Connection at A assumed to be a pin and the cable supports the platform at B
5.2 Free-Body Diagrams

Solution

• Direction of the cable and average dimensions of the platform are listed and center of gravity has been determined
• Free-Body Diagram
5.2 Free-Body Diagrams

Solution

• Platform’s weight = 200(9.81) = 1962N
• Force components Ax and Ay along with the cable force T represent the reactions that both pins and cables exert on the platform
• Half of the cables magnitudes is developed at A and half developed at B
5.2 Free-Body Diagrams

Example 5.5

The free-body diagram of each object is

drawn. Carefully study each solution and

5.3 Equations of Equilibrium
• For equilibrium of a rigid body in 2D,

∑Fx = 0; ∑Fy = 0; ∑MO = 0

• ∑Fx and ∑Fy represent the algebraic sums of the x and y components of all the forces acting on the body
• ∑MO represents the algebraic sum of the couple moments and moments of the force components about an axis perpendicular to x-y plane and passing through arbitrary point O, which may lie on or off the body
5.3 Equations of Equilibrium

Alternative Sets of Equilibrium Equations

• For coplanar equilibrium problems, ∑Fx = 0; ∑Fy = 0; ∑MO = 0 can be used
• Two alternative sets of three independent equilibrium equations may also be used

∑Fa = 0; ∑MA = 0; ∑MB = 0

When applying these equations, it is required that a line passing through points A and B is not perpendicular to the a axis

5.3 Equations of Equilibrium

Alternative Sets of Equilibrium Equations

• Consider FBD of an arbitrarily shaped body
• All the forces on FBD may be

replaced by an equivalent

resultant force

FR = ∑F acting at point A and a

resultant moment MRA = ∑MA

• If ∑MA = 0 is satisfied, MRA = 0
5.3 Equations of Equilibrium

Alternative Sets of Equilibrium Equations

• If FR satisfies ∑Fa= 0, there is no

component along the a axis and

its line of axis is perpendicular

to the a axis

• If ∑MB = 0 where B does not lies

on the line of action of FR, FR = 0

• Since ∑F = 0 and ∑MA = 0, the

body is in equilibrium

5.3 Equations of Equilibrium

Alternative Sets of Equilibrium Equations

• A second set of alternative equations is

∑MA = 0; ∑MB = 0; ∑MC = 0

• Points A, B and C do not lie on the

same line

• Consider FBD, if If ∑MA = 0, MRA = 0
• ∑MA = 0 is satisfied if line of action of FR passes through point B
• ∑MC = 0 where C does not lie on line AB
• FR = 0 and the body is in equilibrium
5.3 Equations of Equilibrium

Procedure for Analysis

Free-Body Diagram

• Establish the x, y, z coordinates axes in any suitable orientation
• Draw an outlined shape of the body
• Show all the forces and couple moments acting on the body
• Label all the loadings and specify their directions relative to the x, y axes
5.3 Equations of Equilibrium

Procedure for Analysis

Free-Body Diagram

• The sense of a force or couple moment having an unknown magnitude but known line of action can be assumed
• Indicate the dimensions of the body necessary for computing the moments of forces
5.3 Equations of Equilibrium

Procedure for Analysis

Equations of Equilibrium

• Apply the moment equation of equilibrium ∑MO = 0 about a point O that lies on the intersection of the lines of action of the two unknown forces
• The moments of these unknowns are zero about O and a direct solution the third unknown can be obtained
5.3 Equations of Equilibrium

Procedure for Analysis

Equations of Equilibrium

• When applying the force equilibrium ∑Fx = 0 and ∑Fy = 0, orient the x and y axes along the lines that will provide the simplest resolution of the forces into their x and y components
• If the solution yields a negative result scalar, the sense is opposite to that was assumed on the FBD
5.3 Equations of Equilibrium

Example 5.6

Determine the horizontal and vertical

components of reaction for the beam loaded.

Neglect the weight of the beam in the

calculations.

5.3 Equations of Equilibrium

Solution

FBD

• 600N force is represented by its x and y components
• 200N force acts on the beam at B and is

independent of the

force components

Bx and By, which

represent the effect of

the pin on the beam

5.3 Equations of Equilibrium

Solution

Equations of Equilibrium

• A direct solution of Ay can be obtained by applying ∑MB = 0 about point B
• Forces 200N, Bx and By all create zero moment about B
5.3 Equations of Equilibrium

Example 5.7

The cord supports a force of 500N and wraps over

the frictionless pulley. Determine the tension in the

cord at C and the horizontal

and vertical components at

pin A.

5.3 Equations of Equilibrium

Solution

FBD of the cord and pulley

• Principle of action: equal but opposite reaction observed in the FBD
• Cord exerts an unknown load

distribution p along part of

the pulley’s surface

• Pulley exerts an equal but

opposite effect on the cord

5.3 Equations of Equilibrium

Solution

FBD of the cord and pulley

• Easier to combine the FBD of the pulley and

contracting portion of the cord so that the

distributed load becomes internal to the system

and is eliminated from the

analysis

5.3 Equations of Equilibrium

Solution

Equations of Equilibrium

Tension remains constant as cord

passes over the pulley (true for

any angle at which the cord is

directed and for any radius of

the pulley

5.3 Equations of Equilibrium

Example 5.8

The link is pin-connected at a and rest a

smooth support at B. Compute the horizontal

and vertical components of reactions at pin A

5.3 Equations of Equilibrium

Solution

FBD

• Reaction NB is perpendicular to the link at B
• Horizontal and vertical

components of reaction

are represented at A

5.3 Equations of Equilibrium

Solution

Equations of Equilibrium

5.3 Equations of Equilibrium

Example 5.9

The box wrench is used to tighten the bolt at

A. If the wrench does not turn when the load

is applied to the handle, determine the

torque or moment applied to the bolt and the

force of the wrench on the bolt.

5.3 Equations of Equilibrium

Solution

FBD

• Bolt acts as a “fixed support” it exerts force components Ax and Ay and a torque MA on the wrench at A
5.3 Equations of Equilibrium

Solution

Equations of Equilibrium

5.3 Equations of Equilibrium

Solution

• Point A was chosen for summing the moments as the lines of action of the unknown forces Ax and Ay pass through this point and these forces are not included in the moment summation
• MA must be included
• Couple moment MA is a free vector and represents the twisting resistance of the bolt on the wrench
5.3 Equations of Equilibrium

Solution

• By Newton’s third law, the wrench exerts an equal but opposite moment or torque on the bolt
• For resultant force on the wrench,
• For directional sense,
• FA acts in the opposite direction on the bolt
5.3 Equations of Equilibrium

Example 5.10

Placement of concrete from the

truck is accomplished using the

chute. Determine the force that the

hydraulic cylinder and the truck

frame exert on the chute to hold it in

position. The chute and the wet

concrete contained along its length

have a uniform weight of 560N/m.

5.3 Equations of Equilibrium

Solution

• Idealized model of the chute
• Assume chute is pin connected to the frame at A and the hydraulic cylinder BC acts as a short link
5.3 Equations of Equilibrium

Solution

FBD

• Since chute has a length of 4m, total supported weight is (560N/m)(4m) = 2240N, which is assumed to act at its midpoint, G
• The hydraulic cylinder exerts a horizontal force FBC on the chute

Equations of Equilibrium

• A direct solution of FBC is obtained by the summation about the pin at A
5.3 Equations of Equilibrium

Example 5.11

The uniform smooth rod is subjected to a force and

couple moment. If the rod is supported at A by a

smooth wall and at B and C either at the top or

bottom by rollers, determine

the reactions at these supports.

Neglect the weight of the rod.

5.3 Equations of Equilibrium

Solution

FBD

• All the support reactions act normal to the surface of contact since the contracting surfaces are smooth
• Reactions at B and C are

acting in the positive y’

direction

• Assume only the rollers

located on the bottom of

the rod are used for support

5.3 Equations of Equilibrium

Solution

Equations of Equilibrium

5.3 Equations of Equilibrium

Solution

• Note that the line of action of the force component passes through point A and this force is not included in the moment equation
• Since By’ is negative scalar, the sense of By’ is opposite to shown in the FBD
5.3 Equations of Equilibrium

Solution

• Top roller at B serves as the support rather than the bottom one
5.3 Equations of Equilibrium

Example 5.12

The uniform truck ramp has a weight of

1600N ( ≈ 160kg ) and is pinned to the body

of the truck at each end and held in position

by two side cables.

Determine the tension

in the cables.

5.3 Equations of Equilibrium

Solution

• Idealized model of the ramp
• Center of gravity located at the midpoint since the ramp is approximately uniform

FBD of the Ramp

5.3 Equations of Equilibrium

Solution

Equations of Equilibrium

By the principle of transmissibility, locate T at C

5.3 Equations of Equilibrium

Solution

Since there are two cables supporting the

ramp,

T’ = T/2 = 2992.5N

5.4 Two- and Three-Force Members
• Simplify some equilibrium problems by recognizing embers that are subjected top only 2 or 3 forces

Two-Force Members

• When a member is subject to

no couple moments and forces

are applied at only two points

on a member, the member is

called a two-force member

5.4 Two- and Three-Force Members

Two-Force Members

Example

• Forces at A and B are summed to obtain their respective resultants FA and FB
• These two forces will maintain translational and force equilibrium provided FA is of equal magnitude and opposite direction to FB
• Line of action of both forces is known and passes through A and B
5.4 Two- and Three-Force Members

Two-Force Members

• Hence, only the force magnitude must be determined or stated
• Other examples of the two-force members held in equilibrium are shown in the figures to the right
5.4 Two- and Three-Force Members

Three-Force Members

• If a member is subjected to only three forces, it is necessary that the forces be either concurrent or parallel for the member to be in equilibrium
• To show the concurrency

requirement, consider a body

with any two of the three forces

acting on it, to have line of

actions that intersect at point O

5.4 Two- and Three-Force Members

Three-Force Members

• To satisfy moment equilibrium about O, the third force must also pass through O, which then makes the force concurrent
• If two of the three forces parallel,

the point of currency O, is

considered at “infinity”

• Third force must parallel to

the other two forces to insect

at this “point”

5.4 Two- and Three-Force Members
• Bucket link AB on the back hoe is a typical example of a two-force member since it is pin connected at its end provided its weight is neglected, no other force acts on this member
• The hydraulic cylinder is pin connected at its ends, being a two-force member
5.4 Two- and Three-Force Members
• The boom ABD is subjected to the weight of the suspended motor at D, the forces of the hydraulic cylinder at B, and the force of the pin at A. If the boom’s weight is neglected, it is a three-force member
• The dump bed of the truck operates by extending the hydraulic cylinder AB. If the weight of AB is neglected, it is a two-force member since it is pin-connected at its end points
5.4 Two- and Three-Force Members

Example 5.13

The lever ABC is pin-supported

at A and connected to a short

link BD. If the weight of the

members are negligible,

determine the force of the pin

on the lever at A.

5.4 Two- and Three-Force Members

Solution

FBD

• Short link BD is a two-force member, so the resultant forces at pins D and B must be equal, opposite and collinear
• Magnitude of the force is unknown but line of action known as it passes through B and D
• Lever ABC is a three-force member
5.4 Two- and Three-Force Members

Solution

FBD

• For moment equilibrium, three non-parallel forces acting on it must be concurrent at O
• Force F on the lever at B is equal but opposite to the force F acting at B on the link
• Distance CO must be 0.5m since lines of action of F and the 400N force are known
5.4 Two- and Three-Force Members

Solution

Equations of Equilibrium

Solving,

5.5 Equilibrium in Three Dimensions (FBD)
• Consider types of reaction that can occur at the supports

Support Reactions

• Important to recognize the symbols used to represent each of these supports and to clearly understand how the forces and couple moments are developed by each support
• As in 2D, a force that is developed by a support that restricts the translation of the attached member
5.5 Equilibrium in Three Dimensions (FBD)

Support Reactions

• A couple moment is developed when rotation of the attached member is prevented

Example

• The ball and socket joint prevents any translation of connecting member; therefore, a force must act on the member at the point of connection
• This force have unknown magnitude Fx, Fy and Fz
• Magnitude of the force is given by
5.5 Equilibrium in Three Dimensions (FBD)

Support Reactions

• The force’s orientation is defined by the coordinate angles α, β and γ
• Since connecting member is allow to rotate freely about any axis, no couple moment is resisted by a ball and socket joint
• Single bearing supports, single pin and single hinge are shown to support both force and couple moment components
5.5 Equilibrium in Three Dimensions (FBD)

Support Reactions

• However, if these supports are used with other bearings, pins or hinges to hold a rigid body in equilibrium and the supports are properly aligned when connected to the body, the force reactions at these supports alone may be adequate for supporting the body
• Couple moments become redundant and not shown on the FBD
5.5 Equilibrium in Three Dimensions (FBD)
• Ball and socket joint provides a connection for the housing of an earth grader to its frame
• Journal bearing supports the end of the shaft
5.5 Equilibrium in Three Dimensions (FBD)
• Thrust bearing is used to support the drive shaft on the machine
• Pin is used to support the end of the strut used on a tractor
5.5 Equilibrium in Three Dimensions (FBD)

Example 5.14

Several examples of objects along with their

associated free-body diagrams are shown. In

all cases, the x, y and z axes are established

and the unknown reaction components are

indicated in the positive sense. The weight of

the objects is neglected.

5.6 Equations of Equilibrium

Vector Equations of Equilibrium

• For two conditions for equilibrium of a rigid body in vector form,

∑F = 0

∑MO = 0

where ∑F is the vector sum of all the external forces acting on the body and ∑MO is the sum of the couple moments and the moments of all the forces about any point O located either on or off the body

5.6 Equations of Equilibrium

Scalar Equations of Equilibrium

• If all the applied external forces and couple moments are expressed in Cartesian vector form

∑F = ∑Fxi + ∑Fyj + ∑Fzk = 0

∑MO = ∑Mxi + ∑Myj + ∑Mzk = 0

• i, j and k components are independent from one another
5.6 Equations of Equilibrium

Scalar Equations of Equilibrium

• ∑Fx = 0, ∑Fy = 0, ∑Fz = 0 shows that the sum of the external force components acting in the x, y and z directions must be zero
• ∑Mx = 0, ∑My = 0, ∑Mz = 0 shows that the sum of the moment components about the x, y and z axes to be zero
5.7 Constraints for a Rigid Body
• To ensure the equilibrium of a rigid body, it is necessary to satisfy the equations equilibrium and have the body properly held or constrained by its supports

Redundant Constraints

• More support than needed for equilibrium
• Statically indeterminate: more unknown loadings on the body than equations of equilibrium available for their solution
5.7 Constraints for a Rigid Body

Redundant Constraints

Example

• For the 2D and 3D problems, both are statically indeterminate because of additional supports reactions
• In 2D, there are 5 unknowns but 3 equilibrium equations can be drawn
5.7 Constraints for a Rigid Body

Redundant Constraints

Example

• In 3D, there are 8 unknowns but 6 equilibrium equations can be drawn

involving the physical

properties of the body

are needed to solve

indeterminate problems

5.7 Constraints for a Rigid Body

Improper Constraints

• Instability of the body caused by the improper constraining by the supports
• In 3D, improper constraining occur when the support reactions all intersect a common axis
• In 2D, this axis is perpendicular to the plane of the forces and appear as a point
• When all reactive forces are concurrent at this point, the body is improperly constrained
5.7 Constraints for a Rigid Body

Improper Constraints

Example

• From FBD, summation of moments about the x axis will not be equal to zero, thus rotation occur
• In both cases,

impossible to

solve completely

for the unknowns

5.7 Constraints for a Rigid Body

Improper Constraints

• Instability of the body also can be caused by the parallel reactive forces

Example

• Summation of

forces along the

x axis will not be

equal to zero

5.7 Constraints for a Rigid Body

Improper Constraints

• Instability of the body also can be caused when a body have fewer reactive forces than the equations of equilibrium that must be satisfied
• The body become partially constrained

Example

• If O is a point not located on line AB, loading condition and equations of equilibrium are not satisfied
5.7 Constraints for a Rigid Body

Improper Constraints

• Proper constraining requires

- lines of action of the reactive forces do not insect points on a common axis

- the reactive forces must not be all parallel to one another

• When the minimum number of reactive forces is needed to properly constrain the body, the problem is statically determinate and equations of equilibrium can be used for solving
5.7 Constraints for a Rigid Body

Procedure for Analysis

Free Body Diagram

• Draw an outlined shape of the body
• Show all the forces and couple moments acting on the body
• Establish the x, y, z axes at a convenient point and orient the axes so that they are parallel to as many external forces and moments as possible
• Label all the loadings and specify their directions relative to the x, y and z axes
5.7 Constraints for a Rigid Body

Procedure for Analysis

Free Body Diagram

• In general, show all the unknown components having a positive sense along the x, y and z axes if the sense cannot be determined
• Indicate the dimensions of the body necessary for computing the moments of forces
5.7 Constraints for a Rigid Body

Procedure for Analysis

Equations of Equilibrium

• If the x, y, z force and moment components seem easy to determine, then apply the six scalar equations of equilibrium,; otherwise, use the vector equations
• It is not necessary that the set of axes chosen for force summation coincide with the set of axes chosen for moment summation
• Any set of nonorthogonal axes may be chosen for this purpose
5.7 Constraints for a Rigid Body

Procedure for Analysis

Equations of Equilibrium

• Choose the direction of an axis for moment summation such that it insects the lines of action of as many unknown forces as possible
• In this way, the moments of forces passing through points on this axis and forces which are parallel to the axis will then be zero
• If the solution yields a negative scalar, the sense is opposite to that was assumed
5.7 Constraints for a Rigid Body

Example 5.15

The homogenous plate has a mass of 100kg

and is subjected to a force and couple

moment along its edges. If it is supported in

the horizontal plane by means of a roller at

A, a ball and socket joint

at N, and a cord at C,

determine the components

of reactions at the supports.

5.7 Constraints for a Rigid Body

Solution

FBD

• Five unknown reactions acting on the plate
• Each reaction assumed to act in a positive coordinate direction
5.7 Constraints for a Rigid Body

Solution

Equations of Equilibrium

• Moment of a force about an axis is equal to the product of the force magnitude and the perpendicular distance from line of action of the force to the axis
• Sense of moment determined from right-hand rule
5.7 Constraints for a Rigid Body

Solution

• Components of force at B can be eliminated if x’, y’ and z’ axes are used
5.7 Constraints for a Rigid Body

Solution

Solving,

Az = 790N Bz = -217N TC = 707N

• The negative sign indicates Bz acts downward
• The plate is partially constrained since the supports cannot prevent it from turning about the z axis if a force is applied in the x-y plane
5.7 Constraints for a Rigid Body

Example 5.16

The windlass is supported by a thrust

bearing at A and a smooth journal bearing at

B, which are properly aligned on the shaft.

Determine the magnitude of the vertical force

P that must be applied to the

handle to maintain equilibrium

of the 100kg bucket. Also,

calculate the reactions at the bearings.

5.7 Constraints for a Rigid Body

Solution

FBD

• Since the bearings at A and B are aligned correctly, only force reactions occur at these supports
5.7 Constraints for a Rigid Body

Solution

Equations of Equilibrium

5.7 Constraints for a Rigid Body

Solution

Equations of Equilibrium

5.7 Constraints for a Rigid Body

Example 5.17

Determine the tension in cables BC and BD

and the reactions at the ball and socket joint

A for the mast.

5.7 Constraints for a Rigid Body

Solution

FBD

• Five unknown force magnitudes
5.7 Constraints for a Rigid Body

Solution

Equations of Equilibrium

5.7 Constraints for a Rigid Body

Solution

Equations of Equilibrium

5.7 Constraints for a Rigid Body

Example 5.18

Rod AB is subjected to the 200N force.

Determine the reactions at the ball and

socket joint A and the

tension in cables BD

and BE.

5.7 Constraints for a Rigid Body

Solution

Equations of Equilibrium

5.7 Constraints for a Rigid Body

Solution

Equations of Equilibrium

Since rC = 1/2rB,

5.7 Constraints for a Rigid Body

Example 5.19

The bent rod is supported at A by a

journal bearing, at D by a ball and

socket joint, and at B by means of

cable BC. Using only one equilibrium

equation, obtain a direct solution for

the tension in cable BC. The bearing at

A is capable of exerting force

components only in the z and y

directions since it is properly aligned.

5.7 Constraints for a Rigid Body

Solution

FBD

• Six unknown
• Three force components caused by ball and socket joint
• Two caused by bearing
• One caused by cable
5.7 Constraints for a Rigid Body

Solution

Equations of Equilibrium

• Direction of the axis is defined by the unit vector
5.7 Constraints for a Rigid Body

Solution

Equations of Equilibrium

Chapter Summary

Free-Body Diagram

• Draw a FBD, an outline of the body which shows all the forces and couple moments that act on the body
• A support will exert a force on the body in a particular direction if it prevents translation of the body in that direction
• A support will exert a couple moment on a body is it prevents rotation
Chapter Summary

Free-Body Diagram

• Angles are used to resolved forces and dimensions used to take moments of the forces
• Both must be shown on the FBD
Chapter Summary

Two Dimensions

• ∑Fx = 0; ∑Fy = 0; ∑MO = 0 can be applied when solving 2D problems
• For most direct solution, try summing the forces along an axis that will eliminate as many unknown forces as possible
Chapter Summary

Three Dimensions

• Use Cartesian vector analysis when applying equations of equilibrium
• Express each known and unknown force and couple moment shown on the FBD as a Cartesian vector
• Set the force summation equal to zero
• Take moments about point O that lies on the line of action of as many unknown components as possible
Chapter Summary

Three Dimensions

• From point O, direct position vectors to each force, then use the cross product to determine the moment of each force
• The six scalar equations of equilibrium are established by setting i, j and k components of these force and moment sums equal to zero